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2x^{2}+2x-73=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-73\right)}}{2\times 2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-2±\sqrt{4-4\times 2\left(-73\right)}}{2\times 2}
Square 2.
x=\frac{-2±\sqrt{4-8\left(-73\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-2±\sqrt{4+584}}{2\times 2}
Multiply -8 times -73.
x=\frac{-2±\sqrt{588}}{2\times 2}
Add 4 to 584.
x=\frac{-2±14\sqrt{3}}{2\times 2}
Take the square root of 588.
x=\frac{-2±14\sqrt{3}}{4}
Multiply 2 times 2.
x=\frac{14\sqrt{3}-2}{4}
Now solve the equation x=\frac{-2±14\sqrt{3}}{4} when ± is plus. Add -2 to 14\sqrt{3}.
x=\frac{7\sqrt{3}-1}{2}
Divide -2+14\sqrt{3} by 4.
x=\frac{-14\sqrt{3}-2}{4}
Now solve the equation x=\frac{-2±14\sqrt{3}}{4} when ± is minus. Subtract 14\sqrt{3} from -2.
x=\frac{-7\sqrt{3}-1}{2}
Divide -2-14\sqrt{3} by 4.
2x^{2}+2x-73=2\left(x-\frac{7\sqrt{3}-1}{2}\right)\left(x-\frac{-7\sqrt{3}-1}{2}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+7\sqrt{3}}{2} for x_{1} and \frac{-1-7\sqrt{3}}{2} for x_{2}.
x ^ 2 +1x -\frac{73}{2} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 2
r + s = -1 rs = -\frac{73}{2}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = -\frac{1}{2} - u s = -\frac{1}{2} + u
Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -\frac{73}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{73}{2}
\frac{1}{4} - u^2 = -\frac{73}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{73}{2}-\frac{1}{4} = -\frac{147}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{147}{4} u = \pm\sqrt{\frac{147}{4}} = \pm \frac{\sqrt{147}}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{1}{2} - \frac{\sqrt{147}}{2} = -6.562 s = -\frac{1}{2} + \frac{\sqrt{147}}{2} = 5.562
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.